Question

Construct∆PQR in which L Q=70° LR=80° and PQ+PR=8.5
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Answer 1

Question :-

Construct a ∆PQR in which LQ=70° LR=80° and PQ+PR=8.5

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Solution :-

☆To Construct :-

Construct a ∆PQR

☆Given Information :-

$\boxed{ \bf \red{\angle Q = 70 \degree}} \\ \boxed{\bf \red{\angle R = 80 \degree}} \\ \boxed{\bf \red{PQ + PR = 8.5 \: cm}}$

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Result :-

Please refer to the image attached above for the construction of the triangle.

Steps :-

Step 1: Draw a line segment AB (PQ+QR+PR=9.5 cm ) = 9.5cm.

Step 2: Draw an angle of 70° at A and an angle of 80° at B.

Step 3: Bisect ∠A and ∠B. Let the bisectors of ∠A and ∠B intersect in P.

Step 4: Draw perpendicular bisector DE of AP which intersect AB in Q and FG of AB which intersect AB in R.

Step 5: Join PQ and PR.

∴ ΔPQR is the required triangle.

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Answer 2

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Question ⤵

• Construct∆PQR in which L Q=70° LR=80° and PQ+PR=8.5

Answer ⤵

Steps of construction:

• Draw seg AB of 8.5cm length.

• Draw a ray making angle of 35 ∘ at point A.

• Draw another ray making an angle of 40∘ at point B.

• Name the point of intersection of the two rays as P.

• Draw the perpendicular bisector of seg AP and segBP.

Name the points as Q and R respectively where the perpendicular bisectors intersect line AB.

• Draw seg PQ and seg PR.

• Therefore, ΔPQR is the required triangle.

Then ⬇

• Your Construction of ∆PQR in which L Q=70° LR=80° and PQ+PR=8.5 Has been completed

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$\huge\fbox\pink{Thank you :)}$